# Primitive of Arcsecant of x over a/Formulation 1

## Theorem

$\ds \int \arcsec \frac x a \rd x = \begin {cases} x \arcsec \dfrac x a - a \map \ln {x + \sqrt {x^2 - a^2} } + C & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\ x \arcsec \dfrac x a + a \map \ln {x + \sqrt {x^2 - a^2} } + C & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\ \end {cases}$

## Proof

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

 $\ds u$ $=$ $\ds \arcsec \frac x a$ $\ds \leadsto \ \$ $\ds \frac {\d u} {\d x}$ $=$ $\ds \begin {cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\ \dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\ \end {cases}$ Derivative of $\arcsec \dfrac x a$

and let:

 $\ds \frac {\d v} {\d x}$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds v$ $=$ $\ds x$ Primitive of Constant

First let $\arcsec \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.

Then:

 $\ds \int \arcsec \frac x a \rd x$ $=$ $\ds x \arcsec \frac x a - \int x \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x + C$ Integration by Parts $\ds$ $=$ $\ds x \arcsec \frac x a - a \int \frac {\d x} {\sqrt {x^2 - a^2} } + C$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds x \arcsec \frac x a - a \map \ln {x + \sqrt {x^2 - a^2} } + C$ Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$

Similarly, let $\arcsec \dfrac x a$ be in the interval $\openint {\dfrac \pi 2} \pi$.

Then:

 $\ds \int \arcsec \frac x a \rd x$ $=$ $\ds x \arcsec \frac x a - \int x \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C$ Integration by Parts $\ds$ $=$ $\ds x \arcsec \frac x a + a \int \frac {\d x} {\sqrt {x^2 - a^2} } + C$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds x \arcsec \frac x a + a \map \ln {x + \sqrt {x^2 - a^2} } + C$ Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$

$\blacksquare$